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24-MMP-A5 Surface Mining Methods and Design · December 2018

Question 26 of 27

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper
Paper: Surface Mining Methods and Design (09-MMP-A5), National Exam, December 2018 — 20 pages, compulsory Question 1 (40 marks, parts 1.1–1.8) plus THREE of five optional Questions 2–6 (20 marks each) normally constitute a complete paper. As a study resource, this solution answers Question 1 in full AND all five optional Questions 2–6.

Reference texts: Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design (3rd ed.) — truck-shovel match factor, dragline stripping geometry, capital cost indexes, open-pit scheduling; SME Mining Engineering Handbook (3rd ed.) — equipment costing, mine dewatering, cost-index escalation.

Question 6.1 in-pit sump pump dewatering (11 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Total static lift, sump (1650 m) to crest (1740 m) = 90 m; pump 1 at 1650 m (submersible), pump 2 at 1675 m (tandem, in series with pump 1); friction negligible. Pump characteristic curves (read from Fig. 6.1, l/s vs. ft of head): 433 HT passes through (0, 180 ft), (100, 115 ft), (150, 65 ft); 431 MT passes through (0, 125 ft), (100, 100 ft), (200, 60 ft).

Given data — pump characteristic curve points (Fig. 6.1)
PumpQ=0 l/sQ=100 l/sHigher-Q point
433, HT (high head)180 ft115 ft65 ft @ 150 l/s
431, MT (high volume)125 ft100 ft60 ft @ 200 l/s

Find. Best pump-pair configuration, resulting flow, per-pump inlet/outlet pressure, required pipe rating, and check-valve implications.

Approach. Since the two pumps operate strictly in SERIES on one continuous, low-friction pipeline, their heads ADD at the common operating flow Q; fit each curve through its three given points, then solve for the Q where the combined head equals the 90 m (295.3 ft) total static lift, for each candidate pairing.

  1. Fit the two curves. A quadratic H=a+bQ+cQ² through each set of three points gives H HT(Q)=180−0.4167Q−0.002333Q² and H MT(Q)=125−0.175Q−0.00075Q².
  2. Convert the required lift. $$H_{req} = \dfrac{90\text{ m}}{0.3048\text{ m/ft}} = 295.3\text{ ft}$$
  3. Test both-MT. Maximum possible combined head (at Q=0, the best case) is 2×125=250 ft — LESS than 295.3 ft required, so two MT pumps in series CANNOT reach the crest at ANY flow. Both-MT is infeasible.
  4. Solve both-HT. $$2\,H_{HT}(Q) = 295.3 \;\Rightarrow\; 0.004667Q^2+0.8333Q-65.28=0$$ Solving the quadratic: $$\boxed{Q \approx 58.5\text{ l/s}}$$ — comfortably within the plotted 0–170 l/s HT curve range.
  5. Check the mixed HT+MT pairing. Solving H HT(Q)+H MT(Q)=295.3 gives Q≈15.2 l/s — feasible but far LOWER flow than the both-HT pairing, so it is dominated.

6.1.1 — configuration choice. $$\boxed{\text{Use BOTH pumps in the high-head (433, HT) configuration}}$$ Two MT pumps cannot reach the required 295 ft combined head at any flow; a mixed HT+MT pair reaches the required head only at a much lower flow (≈15 l/s) than two HT pumps (≈58.5 l/s) — both-HT is the only configuration that both meets the lift and maximises discharge.

6.1.2 — maximum discharge. $$\boxed{Q_{max} \approx 58.5\text{ l/s}}$$ at the crest, using both pumps in the high-head configuration.

6.1.3 — inlet/outlet pressures. Each identical HT pump contributes exactly half the total lift at this flow: $$H_{each} = H_{HT}(58.5\text{ l/s}) = 147.6\text{ ft} = 45.0\text{ m}$$

PointPressure (m water)Pressure (psi, 1 m≈1.41 psi)
Pump 1 inlet (sump, submerged suction)≈0 (flooded suction)≈0
Pump 1 outlet (1650 m)45.0 m63.5 psi
Pump 2 inlet (1675 m, tandem) — pump 1’s head less the 25 m static rise 1650→167545.0−25=20.0 m28.2 psi
Pump 2 outlet (1675 m) — pump 2 adds its own 45.0 m20.0+45.0=65.0 m91.7 psi

Consistency check: pump 2’s outlet head (65.0 m) minus the remaining static rise to the crest (1740−1675=65 m) leaves exactly 0 m — atmospheric discharge at the crest, confirming the solved flow rate closes the whole system with zero friction loss, as assumed.

6.1.4 — pipe schedule. The governing pressure is pump 2’s OUTLET, ≈92 psi (≈632 kPa) — applying a conventional ∼1.6× safety margin over peak operating pressure calls for HDPE pipe rated at least 150 psi (∼PN10, SDR17 class); the lower-pressure pump-1-to-pump-2 run (≈63 psi) would technically clear a lighter SDR26/PN6 (∼90 psi) rating, but specifying the SAME PN10/SDR17 pipe for BOTH runs is the practical recommendation for spares commonality and to retain margin if the operating point shifts.

6.1.5 — check valve at the top pump’s outlet. A check valve stops the 65 m water column above it draining back to the sump when pump 2 stops, but this creates three new requirements: (1) pump 2’s motor must be sized for its full SHUT-OFF head at zero flow on every restart (the trapped column means no gradual pressure build-up), so a soft-start or slow-opening arrangement is needed to avoid overload/water-hammer (Question 1.5) on restart; (2) a slow-closing or spring/damped check valve (not a fast-slam swing check) is needed to avoid a hammer surge every time the pump trips; (3) a pressure-relief/bleed valve downstream of the check valve is needed so the trapped column can be safely drained before any maintenance access to the pipe or pump.

ItemResult
6.1.1 configurationboth pumps in high-head (433, HT) configuration
6.1.2 max discharge≈58.5 l/s
6.1.3 pump 1 outlet / pump 2 inlet / pump 2 outlet45.0 m / 20.0 m / 65.0 m
6.1.4 pipe rating≥150 psi HDPE (PN10/SDR17), both runs
6.1.5 check-valve modificationssoft-start motor/valve, damped check valve, downstream bleed/relief valve